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sweet-ann [11.9K]

4 years ago

9

Olivia places her pet frog on a line to observe the frog motion. The line is divided into sections that measure 1 centimeter eac

h. the frog begins at 0, moves 18 centimeters forward, moves 6 centimeters backwards, and then 12 centimeters backward. What is the frog's displacement?

2 answers:

Goshia [24]4 years ago

4 0

There is no displacement. The frog is back where it began.

Natali [406]4 years ago

3 0

Answer:

displacement of frog is zero

Explanation:

Displacement of the frog is given as

$d_1$ = 18 cm forward

$d_2$ = 6 cm backwards

$d_3$ = 12 cm backwards

so here we have

$d_1 = 18 cm$

$d_2 = -6 cm$

$d_3 = -12 cm$

so we have

$d = d_1 + d_2 + d_3$

$d = 18 - 6 - 12$

$d = 0$

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3 years ago

Derive equation of motion s=ut+1/2at²

Pavel [41]

Recall the definitions of

• average velocity:

v[ave] = ∆x/∆t = (x[final] - x[initial])/t

Take the initial position to be the origin, so x[initial] = 0, and we simply write x[final] = s. So

v[ave] = s/t

• average acceleration:

a[ave] = ∆v/∆t = (v[final] - v[initial])/t

Assume acceleration is constant (a[ave] = a). Let v[initial] = u and v[final] = v, so that

a = (v - u)/t

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Then

v[ave] = s/t = (v + u)/2 ⇒ s = (v + u) t/2

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2 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

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